Team (First Place) Record Pts Last Week
1 Colorado College (30) 23-4-5 589 1
2 Cornell (10) 20-4-1 553 3
3 Maine 22-5-5 505 2
4 New Hampshire 19-7-4 466 6
5 Boston College 19-8-3 406 4
6 Minnesota 17-7-7 384 7
7 North Dakota 22-6-4 375 5
8 Ferris State 23-8-1 313 9
9 Boston University 20-10-2 268 11
10 Michigan 21-8-1 256 8
11 Ohio State 21-7-3 190 10
12 MSU-Mankato 14-7-9 131 12
13 Denver 18-9-5 121 14
14 St. Cloud State 14-11-3 85 15
15 Harvard 16-8-1 78 13
EDIT: forgot to include these...
Others receiving votes: Providence 35,
Michigan State 19, Yale 18, Minnesota-Duluth 5,
Dartmouth 2, Miami 1
This is exciting. Polls don't mean anything, especially at this point of the year, but following them can be a game itself. 10 first place votes and being closer to #1 than we are to #3...wow.
I still like the fact that we only have 5 non-wins. Next closest on the list...CC, FSU, Mich, and Harvard with 9.
This is what you call "arriving". This is probably the first time in history anybody ever dropped a point to an ECAC team (and an unranked one at that) and went up in a poll.
Ouch... Harvard whoops an unranked team, loses a nailbiter to #2, and gets dropped two spots.
Not that surprising: Harvard fell behind Denver, who took 3 points off of then #5 (now #7) NoDak and SCSU, who split with #1 CC.
My initial response was HOLY F#$%!! ......... not about the ranking but about the # of #1 votes ... 10!!!! WOW!
Likely vote distribution:
Team 1st 2nd 3rd Points
CC 30 9 1 589
Cornell 10 13 17 553
I think that's the only way for CC to get 589 with 40 votes. There are other combinations for our ranking and maybe it's likely that we had some lower votes. But every #4 vote would have to be compensated for by an additional #2.
No point in speculating much as you go further down in the poll, except that Maine has to have some votes for #4 or lower.
how exactly does it work? how many points for a 1st place, for a second place, etc.?
40 voters. 15 points for 1st place vote, 14 for 2nd, 13 for 3rd, etc.
I haven't said anything all year, as I've just been enjoying the season in silence. Now, however...
Dayumm! 2nd place? 10 votes for first? Of course polls are meaningless in terms of how well a team is going to play on the following weekend, but it sure is a good indicator of confidence.
I must admit that I am very excited. Sorry not to have made it to the game on Saturday, but by the time I started looking for tickets, the game had been sold out. Apparently having received numerous calls from the Boston-area Cornell alumni, the Harvard ticket office even went so far as to put a message on their answering maching saying that there were no tickets, not even SRO, left for the Saturday game. =)
Cheers,
Kyle
Some further thought about the numbers leads me to guess that we had more second place votes than 13, with fewer 3rd's and some 4th's to balance out. Wish I could easily back out the vote numbers. Unfortunatley I'm not sure you can solve the Ax=b system for A given x and b, even knowing a constraint u'A=c (given u and c). There might be a regression type approach that would work, but it doesn't jump to mind (esp. considering you have to restrict to integers). And the brute force approach is not reasonable.
I really need to do some real work instead of working on thought problems about ceramic dalmatians...
You have to analyze the sums for all the top 15 and others receiving votes to even slightly have an idea of what number of which votes we received. I am not about to even try. It is nearly impossible.
The magic square problem from hell.... I'm 99.99% positive that there would be a set of points that could be achieved with two different sets of votes; therefore, the problem is not solveable in general. On the other hand, if all the voters voted exactly the same, then the problem is solveable - so there are specific cases which would be solveable. The question is, how could you tell by looking at the scores if it is solveable, or more generally, would there be a way to calculate the number of different voting schemes that would have resulted in those scores? Where's a numbers theorist when you need one?
We know that Miami got one 15th place vote... ::nut::
Does anyone else find the little muscle guy in Adam Doyle's sig extremely disturbing?
Well, I've managed to get part of the problem set up as a regression problem. But it doesn't quite work right (no surprise there): my results currently say that Dartmouth received -183.667 first place votes, which counter their 40 votes for each of 2nd, 3rd, 4th, 7-9th and 12-14th places. Dartmouth is really popular!
(Haven't quite figured out how to mathematically restrict each team to one vote per ballot yet. And the pesky integer thing...)
I think the way to go about trying to solve this would be to set up a three dimensional array of variables, x[i,j,k]. Let i iterate through each ballot/voter, let j iterate through each possible placings (ie 1 - 15), and let k iterate through each team. A total of 40*15*21 = 12,600. A little much for my student edition solver, but not that far out.
The constraints would be:
For each voter-placing [i,j], the sum over k must equal 1. (600 constraints). There might be a way to simplify this, but I'm not entirely sure.
For each team k, the sum over all voters and placings, multiplied by the appropriate weights, is equal to the number of points they got (21 constraints).
Each x[i,j,k] is required to be either 0 or 1.
If you wanted to, you could try to set it up as a simple linear algebra problem, where x is a 12600 entry vector, b is a vector with 621 entries (the first 600 would be 1's, the next 21 would be the various numbers of points received). The A matrix would have 621 rows, and 12600 columns, simply containing the coefficients for each of the constraints.
However, this would not do the trick at all, because of the symmetry between the different voters. Trying to just work it out by hand might be easier.
The goal is to determine all of the unknown votes, right? So that's 14*40 = 560 unknowns.
Of the knowns, we have the sum of each teams' votes, which can be expressed in the form
139 = 14*cc_2 + 13*cc_3 + 12*cc_4 + ... + cc_15
403 = 14*cor_2 + 13*cor_3 + 12*cor_4 + ... + cor_15
:
1 = mu_15.
(We can get rid of the 1st place vote variable, since those are known.)
You can also say that each voter can only cast 1 vote per place. In other words:
40 = cc_2 + cor_2 + me_2 + ... + yu_2
40 = cc_3 + cor_3 + me_3 + ... + yu_2
:
40 = cc_15 + cor_15 + me_15 + ... + mu_15.
There are 21 team equations and 14 ballot equations, so that gives us 35 total. Without further restriction, you can't do it. Now, you could start making up rules like to say each higher-ranked team receives more higher-ranked votes than the teams below it, which isn't entirely realistic but not bad. After all, Cornell probably didn't get any 15th-place votes, and PC probably didn't get any second-place votes. That task is left to the reader.
Now, what pet does the German own?
P.S. Age, when you edit a post with PRE tags, it puts BRs at the start of every line, thus the double spacing. Anything you can do about that?
Actually, you have 21 more equations, since each team can only have a up to 40 votes. Unfortunately, these are pretty much inequalities since we don't know how many votes each team received.
You can set this up as a least squares problem and get a solution. But since it's underconstrained, the linear LS answer doesn't necessarily give you the right answer. In fact the chances of picking the right one are vanishingly small (infinite number of solutions that satisfy the equations). Now, limiting the range of each variable to [0,40] would cut down the possibilities to a smaller infinity, but still wouldn't get you to the answer. this also pretty much eliminates the linear LS approach. Limiting the answers to integers would get you closer, but I'm not sure how to "phrase" that mathematically, unless you're just searching the space.
You are going to regret doing all of this if you aren't finished by the time the polls come out next Monday.
Well, Cornell got no less than 13 and no greater than 28 second place votes. Does that help narrow down the possibilities at all? ::screwy::
Pat Carr '96 wrote:
QuoteP.S. Age, when you edit a post with PRE tags, it puts BRs at the start of every line, thus the double spacing. Anything you can do about that?
I remember wrestling with that problem for a long time when I set eLF up initially and came to the conclusion that I couldn't do much about it. Don't remember what the deal was, but I don't have time to revisit it now, anyway.
PRE is a problematic tag anyway. You'll find different browsers interpreting it in different ways. I would not be shocked to see it defacated, er... well you know, before long.
You mean deprecated? I should think not. It works just fine as long as you don't go specify a non-monospace type in your font preferences for PRE formatting. This particular problem involves line breaks being converted to CR's in the edit window, then back for HTML displaying and the PRE tags not actually being PRE tags. It's just a stylesheet that simulates PRE formatting for reasons I don't want to get into. The edit window can't tell that part of the text in a post is PRE and part isn't without getting into some unsavory presumptions. The default font for phorum is Courier to get around that. I changed it to something more pleasant to look at and adding the PRE tag option. Anyway, it's complicated and has absolutely nothing to do with actual HTML PRE tags (which are fine in their own right).
Naah. We're sharper than that (a little anyway). The point would be to put together a computer code that would solve for the votes given all the known data. I'm not sure it's really possible and I'm not really putting anymore time into it, but that was my original thought (aside from killing some time).
Naturally I meant deprecated. That's what I wrote. And at 1 in the morning too -- not too shabby. :-D
Not that I remember anything else about it, but can't you solve lots of simultaneous equations with relatively simple matrix manipulation? One row per team, 15 columns per row. A given cell value a*x where a = 15, 14, 13, ... and x is the unknown number of votes of that weight. The sum of each column of a's is 40. Hmmm... 15xN unsolved values for N=number of teams receiving votes and ... ok, maybe that doesn't sound too promising.
Can you just fill the matrix all the way out with a few hundred additional rows with all cofficients = 0, or does that also contribute no useful information?
But we need a eLynah distributed computing project! We can solve for votes and have a neato screensaver with Age's pretty pictures and everything.
Plus it is a better use of space cycles than SETI@home..... ;)
::rock::
Simultaneous equations are easy to solve. Unfortunately we have far more variables (15 per team, or 14 since we know first place votes) than equations. So it's an underconstrained problem and not directly solvable.
Greg wrote:
QuoteCan you just fill the matrix all the way out with a few hundred additional rows with all cofficients = 0, or does that also contribute no useful information?
You can't get something for nothing. If you try to add trivial equations to an underdetermined system, the resulting matrix had zero determinant and can't be inverted.
Yeah, I had that instinct, although every once in a while a cutsey little endrun would crop up in my quant methods class where you would effectively add a ton of null data to a series of regression equation and... poof... all of a sudden you got more information.
Mind you, that was going on 14 years ago now, so for all I know... ::screwy::
For the non-math-oriented out there (Grant) I think the essential difficulty is this:
Suppose, at the very bottom of the rankings, you have that both Ithaca College and Tompkins County Community College get 4 points, and you have the rest of the solutions all figured out so that you know they each got two votes, and between them, there is 1 13th place, 2 14th places, and 1 15th place. There is absolutely no way to tell whether IC got the 13th & 15th, and T3C got the 2 14ths or vice versa.
Of course, you can do this for any pair of teams that, for some subset of the votes they got, have the same number of votes with the same number of points. Even adding the constraint that each ballot can only contain at most one of each team doesn't help things, because it could well be that, in the example above, it was a single pair of Ithaca-loving voters that gave points to both IC and T3C, and so the switching within their ballots wouldn't make anything infeasible. So, to know exactly how many of each type of vote each team got, is not generally doable.
Now, just because it's not generally doable doesn't mean that it's never possible. For example, as Adam '03 said, it's obvious that Miami got 1 15th place vote, and no other votes. Similarly, given that Colorado College got exactly 30 1st place votes, I'm pretty sure the breakdown given by KeithK '93 is unique.
Aside from special cases such as those, though, I think the only way to know for sure would be to apply some social engineering, such as flat-out asking USCHO, or doing your own re-poll of the voters.
Hypothetical question. If there was just one more number for each team: the total ballots on which that team was included, would there then be sufficient information?
Aside: the more I look at this (now a whole 3 minutes, up from 2), the more it looks like one of those best-fit multiple node problems. Avaya has 300 national nodes and wants to maximize their screwing of the consumer. What is the most efficient way for them to route their volume? And then you start out with seed values and run some sort of iterative fitting algorithm a few million times until the pct change from case N to N+1 drops below .0000000000000001%.
Of course, this is the kind of thing that smartass Chaos Theorists win Nobels for by proving there are various systems that will never settle down, no matter how many iterations, instead oscillating wildly between random values. But as those guys rarely have offspring, maybe they will iterate themselves out of the picture.
wow, for something that people say means absolutely nothing, you guys are going to an AWFUL lot of trouble to figure out what our votes were. Obviously it doesn't mean nothing. But who really cares how the votes break down? and if you do, why?
and if you say you are doing it for fun, I swear you need to find a better hobby.
To be honest I started trying to figure this out 'cuz I was bored at work. And I find solving mathematical problems - particularly those involving meaningless things like hockey polls - to be strangely intersting and fun. But hey, PhD rocket scientists are supposed to be geeks, right?
Ok, if you guys want another math problem that many have considered and none have solved try this:
Pennsylvania used to have a Super 7 lottery, 80 numbers, you pick 7, then they pull 11 numbers, and you win if your 7 are in their 11. What is the minimum number of tickets that must be bought to guarantee at least 1 winning ticket?
If this crowd can't solve this then I might try submitting it to the math dept. and letting them try.
But where's the obligatory hockey content?
I think it has to be at least (80!7!4!)/(11!69!11!). That's the number you need if you can manage it with no "overlap" between your tickets, i.e., no possibility of the same drawing winning for two of your tickets. But I don't know off the top of my head if that's possible. (The reasoning is that there are 80!/(11!69!) possible drawings, but any given ticket will win the prize for 11!/(7!4!) different drawings.)